Question

How do you solve the quadratic using the quadratic formula given `3x^2-11x=8-14x` over the set of complex numbers?

Answer 1

Solution is `-1/2-sqrt105/6` and `-1/2+sqrt105/6`

Explanation:

For the equation `ax^2+bx+c=0`, the roots are given by `x=(-b+-sqrt(b^2-4ac))/(2a)`. As can be seen type of roots depend upon discriminant `b^2-4ac`. If it is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.

In the equation `3x^2-11x=8-14x` or `3x^2+3x-8=0`, the discriminant is `(3)^2-4×3×(-8)=9+96=105`, which is not a complete square and hence roots are irrational (not complex as mentioned in question).

Roots are `x=(-3+-sqrt105)/(2×3)`

= `(-3+-sqrt105)/6`

= `-1/2+-sqrt105/6`

Hence solution is `-1/2-sqrt105/6` and `-1/2+sqrt105/6`.